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An electron of mass me and a proton of mass mp are accelerated through the same potential difference. The ratio of the de Broglie wavelength associated with anelectron to that associated with proton is

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a

1

b

mp/me

c

me/mp

d

mp/me

answer is D.

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Detailed Solution

If q is the charge on the particle and V the potential difference through which it is accelerated, thenqV=12mv2 or mv=2mqVde Broglie's wavelength,λ=hmv=h2mqV ∴λeλp=mpme

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